# Physics check

A performer, seated on a trapeze, is swinging back and forth with a period of 9.55 s. If she stands up, thus raising the center of mass of the trapeze + performer system by 20.0 cm, what will be the new period of the system? Treat trapeze + performer as a simple pendulum.
(a) ____ s

..this is what i did:

T = 2pi sqrt(2L/3g)
1.64 s = 2pi sqrt(((2)(Xm))/((3)(9.8m/s^2)))
[[rearranged to solve for Xm]]

Xm = (((1.64s/2pi)^2)((3)(9.8m/s^2)))/(2)
Xm = 3307.99 m
[[the center of mass increased 20.0cm]]

Xm2 = 3307.99 m + 0.2 m
Xm2 = 3308.19 m
[[plug into original equation]]

T = 2pi sqrt(2L/3g)
T = 2pi sqrt(((2)(3308.19m))/((3)(9.8m/s^2)))
T = 94.3 s
[[final answer]]

.. did i do it right? .. if not, what am i missing or do wrong? thanks!

No, you did not do it right. The work makes no sense to me.

From T=2PI sqrt (l/g), solve for l in the original equaiton. Your work above is similar, but the wrong equation (where did 2/3 come from, and 1.64?
Now, knowing the length l, add twenty cm. Refigure the period.

Wouldn't you have a clue when you found the length of the pendulum as 3308meters long? That is not reasonable.

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asked by COFFEE

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